find all relative extrema of the function. use the second derivative test where applicable. (if an answer…

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)\n\n( f(x)=x^{3}-6 x^{2}+2 )\n\nrelative maximum ( (x, y)=(quad) )\n\nrelative minimum ( (x, y)=(quad) )

find all relative extrema of the function. use the second derivative test where applicable. (if an answer does not exist, enter dne.)\n\n( f(x)=x^{3}-6 x^{2}+2 )\n\nrelative maximum ( (x, y)=(quad) )\n\nrelative minimum ( (x, y)=(quad) )

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (y = f(x)=x^{3}-6x^{2}+2), the first derivative (f^\prime(x)=3x^{2}-12x). Set (f^\prime(x) = 0), so (3x^{2}-12x=3x(x - 4)=0). Solving (3x(x - 4)=0) gives (x = 0) or (x = 4). These are the critical points.

Step2: Find the second derivative

Differentiate (f^\prime(x)=3x^{2}-12x) with respect to (x). Using the power rule, (f^{\prime\prime}(x)=6x-12).

Step3: Apply the second - derivative test

  • For (x = 0): Substitute (x = 0) into (f^{\prime\prime}(x)), (f^{\prime\prime}(0)=6\times0 - 12=- 12<0). When (x = 0), (y=f(0)=0^{3}-6\times0^{2}+2 = 2). So ((0,2)) is a relative maximum.
  • For (x = 4): Substitute (x = 4) into (f^{\prime\prime}(x)), (f^{\prime\prime}(4)=6\times4-12 = 12>0). When (x = 4), (y=f(4)=4^{3}-6\times4^{2}+2=64 - 96 + 2=-30). So ((4,-30)) is a relative minimum.

Answer:

relative maximum ((x,y)=(0,2)) relative minimum ((x,y)=(4,-30))